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Alie Ibarra C16 #59

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191 changes: 148 additions & 43 deletions linked_list/linked_list.py
Original file line number Diff line number Diff line change
@@ -1,79 +1,166 @@

# Defines a node in the singly linked list
from requests import head


class Node:

def __init__(self, value, next_node = None):
def __init__(self, value, next_node = None, prev_node = None):
self.value = value
self.next = next_node
self.previous = prev_node

# Defines the singly linked list
class LinkedList:
def __init__(self):
self.head = None # keep the head private. Not accessible outside this class
self.head = None # keep the head private. Not accessible outside this class
self.tail = None
Comment on lines +11 to +17

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😃 Nice! Doubly Linked List!


#----------------------
# returns the value in the first node
# returns None if the list is empty
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(1)
# Space Complexity: O(1)
def get_first(self):

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pass

if self.head:
return self.head.value
return None

#----------------------
# method to add a new node with the specific data value in the linked list
# insert the new node at the beginning of the linked list
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(1)
# Space Complexity: O(1)

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def add_first(self, value):
pass

if self.head is None:
self.head = self.tail = Node(value)
else:
new_node = Node(value, self.head)
self.head.previous = new_node
self.head = new_node

#----------------------
# method to find if the linked list contains a node with specified value
# returns true if found, false otherwise
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def search(self, value):

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pass
current = self.head
if current == None:
return False
while current:
if current.value == value:
return True
current = current.next
return False

#----------------------
# method that returns the length of the singly linked list
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def length(self):

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pass
counter = 0
current = self.head

while current != None:
current = current.next
counter +=1
return counter

#----------------------
# method that returns the value at a given index in the linked list
# index count starts at 0
# returns None if there are fewer nodes in the linked list than the index value
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def get_at_index(self, index):

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pass
counter = 0
current = self.head

while current != None:
if counter == index:
return current.value
counter += 1
current = current.next
return None

#----------------------
# method that returns the value of the last node in the linked list
# returns None if the linked list is empty
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def get_last(self):

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✨ However, you could also just access the tail's value since this is a doubly linked list and achieve O(1) time complexity

pass
current = self.head
if current == None:
return None

while current.next != None:
current = current.next
return current.value

#----------------------
# method that inserts a given value as a new last node in the linked list
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def add_last(self, value):

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✨ However, again, since you have a tail node, you could just make the new node the new tail and redirect the pointers of the current tail for an O(1) time complexity solution

pass
if self.head == None:
self.add_first(value)

else:
new_node = Node(value)
current = self.head

while current.next:
current = current.next
current.next = new_node

#----------------------
# method to return the max value in the linked list
# returns the data value and not the node
def find_max(self):

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pass
max_value = 0
if self.head == None:
return None

current = self.head
while current:
if current.value > max_value:
max_value = current.value
current = current.next

return max_value

#----------------------
# method to delete the first node found with specified value
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def delete(self, value):

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pass
current = self.head

if current == None:
return

while current:
if current.value == value:
if current == self.head:
self.head = current.next
self.head.previous = None
return

if current == self.tail:
self.tail = current.previous
self.tail.next = None
return

current.previous.next = current.next
current.next.previous = current.previous
return
current = current.next

#----------------------
# method to print all the values in the linked list
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def visit(self):

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helper_list = []
current = self.head
Expand All @@ -84,34 +171,52 @@ def visit(self):

print(", ".join(helper_list))

#----------------------
# method to reverse the singly linked list
# note: the nodes should be moved and not just the values in the nodes
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def reverse(self):

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pass

## Advanced/ Exercises
if not self.head:
return

current_node = self.head
previous_node = None

while current_node:
next = current_node.next
current_node.next = previous_node
previous_node = current_node
current_node = next

self.head = previous_node


#---------------------- Advanced/ Exercises -----------------------
# returns the value at the middle element in the singly linked list
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def find_middle_value(self):

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pass
mid_value = int(self.length() / 2)
return self.get_at_index(mid_value)

#----------------------
# find the nth node from the end and return its value
# assume indexing starts at 0 while counting to n
# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n)
# Space Complexity: O(1)
def find_nth_from_end(self, n):

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pass
nth_from_end = self.length() - (n+1)
return self.get_at_index(nth_from_end)

#----------------------
# checks if the linked list has a cycle. A cycle exists if any node in the
# linked list links to a node already visited.
# returns true if a cycle is found, false otherwise.
# Time Complexity: ?
# Space Complexity: ?
def has_cycle(self):
pass
return False

# Helper method for tests
# Creates a cycle in the linked list for testing purposes
Expand Down
12 changes: 6 additions & 6 deletions tests/linked_list_test.py
Original file line number Diff line number Diff line change
Expand Up @@ -197,7 +197,7 @@ def test_reverse_will_reverse_five_element_list(list):
for i in range(0, 5):
assert list.get_at_index(i) == i

@pytest.mark.skip(reason="Going Further methods")
#@pytest.mark.skip(reason="Going Further methods")
def test_find_middle_value_returns_middle_element_of_five_element_list(list):
list.add_first(10)
list.add_first(30)
Expand All @@ -206,7 +206,7 @@ def test_find_middle_value_returns_middle_element_of_five_element_list(list):
list.add_first(20)
assert list.find_middle_value() == 50

@pytest.mark.skip(reason="Going Further methods")
#@pytest.mark.skip(reason="Going Further methods")
def test_find_middle_value_returns_element_at_index_two_of_six_element_list(list):
list.add_first(10)
list.add_first(30)
Expand All @@ -216,11 +216,11 @@ def test_find_middle_value_returns_element_at_index_two_of_six_element_list(list
list.add_first(100)
assert list.find_middle_value() == 60

@pytest.mark.skip(reason="Going Further methods")
#@pytest.mark.skip(reason="Going Further methods")
def test_nth_from_n_when_list_is_empty(list):
assert list.find_nth_from_end(3) == None

@pytest.mark.skip(reason="Going Further methods")
#@pytest.mark.skip(reason="Going Further methods")
def test_find_nth_from_n_when_length_less_than_n(list):
list.add_first(5)
list.add_first(4)
Expand All @@ -230,7 +230,7 @@ def test_find_nth_from_n_when_length_less_than_n(list):

assert list.find_nth_from_end(6) == None

@pytest.mark.skip(reason="Going Further methods")
#@pytest.mark.skip(reason="Going Further methods")
def test_find_nth_from_n(list):
list.add_first(1)
list.add_first(2)
Expand All @@ -243,7 +243,7 @@ def test_find_nth_from_n(list):
assert list.find_nth_from_end(3) == 4
assert list.find_nth_from_end(4) == None

@pytest.mark.skip(reason="Going Further methods")
#@pytest.mark.skip(reason="Going Further methods")
def test_has_cycle(list):
assert list.has_cycle() == False

Expand Down